The decomposition of `N_(2)O` into `N_(2)` and O in the presence of gaseous argon follows second order kinetics with
`k=(5.0xx10^(11) L " mol"^(-1)s^(-1))e^(-29000 K//T)`
The activation energy for the reaction `E_(a)` is

Options

A. `5.0xx10^(11)` k cal
B. `29000` k cal
C. 58 k cal
D. `-29000` cal

Explanation

To find the activation energy \( E_a \) for the decomposition of \( N_2O \) into \( N_2 \) and \( O \), we can use the Arrhenius equation:

\[ k = A e^{-\frac{E_a}{RT}} \]

where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

From the given information, we have:

\[ k = (5.0 \times 10^{11} \, \text{L mol}^{-1} \text{s}^{-1}) e^{-\frac{29000 \, K}{T}} \]

### Step 1: Identify the relationship between \( E_a \) and the given equation

From the expression for \( k \), we can see that the term \( e^{-\frac{29000}{T}} \) corresponds to \( e^{-\frac{E_a}{RT}} \). Thus, we can equate:

\[ \frac{E_a}{R} = 29000 \]

### Step 2: Solve for \( E_a \)

To find \( E_a \), we can rearrange the equation:

\[ E_a = 29000 \times R \]

### Step 3: Substitute the value of \( R \)

The universal gas constant \( R \) can be expressed in calories as:

\[ R = 1.987 \, \text{cal K}^{-1} \text{mol}^{-1} \]

Now substituting this value into the equation for \( E_a \):

\[ E_a = 29000 \times 1.987 \, \text{cal K}^{-1} \text{mol}^{-1} \]

### Step 4: Calculate \( E_a \)

Now we perform the multiplication:

\[ E_a = 29000 \times 1.987 \approx 57430 \, \text{cal/mol} \]

### Step 5: Convert to kilocalories

To convert calories to kilocalories, we divide by 1000:

\[ E_a \approx \frac{57430}{1000} \approx 57.43 \, \text{kcal/mol} \]

### Step 6: Round to the nearest whole number

Rounding \( 57.43 \) gives us:

\[ E_a \approx 58 \, \text{kcal/mol} \]

### Conclusion

Thus, the activation energy \( E_a \) for the reaction is approximately:

\[ \boxed{58 \, \text{kcal/mol}} \]